Demand paging with page fault

A demand paging system takes 100 time units to service a page fault if it is not a dirty page and 300 time unit as page fault service time if it is a dirty page. memory access time is 1 time unit.the probability of a page fault is case of page fault,the probability of page being dirty is also p. it is observed that the averege access time is 3 time units.then the value of 'p' is.

Arjun @arjunsinghra
21 Jan 2017 07:55 pm

Average memory access time=p(page fault service time + main memory access time) + (1-p)(main memory access time)

page fault service time=page being dirty(page fault service time)+page not dirty(page fault service time)

page fault service time=p(300)+(1-p)(100)

page fault service time=300p+100-100p

page fault service time=200p+100

Average memory access time=p(200p+100 + 1) + (1-p)(1)

Average memory access time=p(200p+101) + (1-p)

Average memory access time=200p2+100p + 1

200p2+100p -2=0

procedure is right??


Arjun @arjunsinghra
16 Jan 2017 06:42 pm

@sumitverma please see

Sumit Verma @sumitkgp
16 Jan 2017 07:11 pm

@arjunsinghra yes, you are right.

Arjun @arjunsinghra
16 Jan 2017 07:19 pm

 but what should be the answer

Sumit Verma @sumitkgp
16 Jan 2017 07:52 pm

It's 0.01925, What is given?

Arjun @arjunsinghra
16 Jan 2017 08:30 pm

i dont know i am getting same

Rahul Ranjan @rahulranjan
21 Jan 2017 09:22 pm

it's 0.0194.